Hard phsyics problem!?

Question

A 1270 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1860 N crate hangs from the far end of the beam

(a) Calculate the magnitude of the tension in the wire. Newtons

(b) Calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam

Fx = N
Fy = N

http://img107.imageshack.us/img107/9378/p966altcu2.gif

for part A
0=T*sin 50 - 1860N - 1270N - ((T*cos 50)/cos 30)*sin 30
solve for T

but how do i solve to T? and is this correct?

Answer 1

Nothing is moving so the net acceleration everywhere must be 0, which means that both the net force everywhere, and the net torque on the beam must all be 0.

For part A, are you looking at the tension in the wire or in the cable supporting beam? If they were one piece with the "joint" at the end of the beam being a pulley, the tension would be the same, but it looks like the beam is pinned at the wall and the cable and wire are both fastened to the beam. In that case the two tensions are different.

For the hanging weight, the tension on the wire and the force of gravity on the weight must balance. This lets you compute the tension in the wire.

Part b is the harder and more interesting one.

At the point where the cable, the wire and the end of the beam meet, you also have a net force of 0. Note that in addition to the two tensions and the force along the beam, you also have the force of gravity acting on the beam itself.

As for the beam itself, you have forces at both ends. Not only must the sum of the forces on the beam be 0, they torques generated by the forces around the center of rotation must also be 0. (You can choose the center of rotation to be anywhere, but the two obvious choices are the center of the beam and the wall end of the beam. I would choose the latter)

(For linear acceleration, a force F up at one end can be balanced by a force F down at the other, but this would create a net torque.)

Each of these generates a set of equations that you can then solve.

I would start with the torque equation on the beam with the axis at the wall (the three forces are the two tensions and the weight of the beam - you can ignore the wall forces because their R value is 0 so they don't contribute to the torque) That should give you the tension in the cable (you already have the tension in the wire).

With the tension on the cable, you can set up the equations that balance net horizontal and vertical forces on the beam. There are two equations - one for the horizontal and one for the vertical - and two unknowns - the wall forces.